Solve using the quadratic formula: x2 – 5x – 3 = -7?

Question

thanks

Answer 1

Get it to the form: ax^2 + bx + c = 0 by subtracting 7 from both sides.

x^2 - 5x + 4 = 0

where a=1, b=-5, c=4

Plug it into the quadratic equation and you get:

x= ( -b +/- sqrt(b^2 - 4ac) ) / 2a

x = ( 5 +/- sqrt((-5)^2 - 4*1*4) ) / 2(1)

x = ( 5 +/- sqrt(9) ) /2

x = (5 +/- 3) / 2

x = 4, 1

Answer 2

What we have to do here is make the equation equal to 0. So all we have to do is use the change sides change signs method with the -7, and we get:
x^2-5x-3+7=0
x^2-5x+4=0

Now, the quadratic equation is (-b±√b²-4ac)/2a. The equation that's good to use along with this equation to make it easier to get the answer right is ax²+bx+c=0.
Looking at our equation right now, we see that a=1, since 1*x²=x². b=-5 because +(-5)=-5 and c=4 by substitution (as you see, x²-5x+4=0 is directly related to the equation ax²+bx+c=0).
Now, plugging in our variables, we get

(-(-5)±√(-5)²-4(1*4)))/2(1).

Now, we know that -(-5)=5. Also, by using the PEMDAS rule with the square root, we know that (1*4)=4, (-5)²=25 and -4(4)=-16. And, in the denominator, 2(1)=2, therefore, we have:

(5±√25-16))/2.

We know that 25-16=9, so:

(5±√9)/2.

The sqaure root of 9 is three, so we have:

(5±3)/2

So, our values for x are: (5+3)/2 and (5-3)/2. We then simplify and get 8/2 and 2/2. Simplifying further, we get 4 and 1. Therefore, your answer is:

x=4,1

There you go. I'm glad I could help. :)

Answer 3

Although the previous answerers solved the problem, they did not use the quadratic formula.

With the equation in the form A x^2 + Bx + C = 0 the solution is
x = {-b +/- sqrt[(b^2)-4ac]}/2a

x= [5 +/- sqrt(25-16)]/2
x=4, 1

Answer 4

x2 - 5x + 4 = 0
if you factorise it it becomes
(x - 4) (x - 1) = 0
so one of the two terms has to equal 0 soo...
x=4 AND/OR x=1

Answer 5

x² - 5x -3 = -7
x² - 5 + 4 = 0
(x - 4)(x-1) = 0
x = 1 or x = 4

Answer 6

(x-1)(x-4)=0

Add 7 to both sides, and factor.
x=1 or 4